Steelers OLB Alex Highsmith Wins AFC Defensive Player of Week

Pittsburgh Steelers outside linebacker Alex Highsmith has been named the AFC Defensive Player of the Week for Week 10 of the 2022 NFL season, the league announced on Wednesday.

Highsmith had two sacks, one forced fumble and five total tackles in the Steelers’ 20-10 win over the New Orleans Saints. It’s the first time Highsmith has won the award in his three-year career. He is the second Steelers player to win the award this season, after safety Minkah Fitzpatrick won it in Week 1.

It was Highsmith’s third multi-sack game of the season, after recording three sacks against the Bengals in Week 1 and 1.5 against the Cleveland Browns in Week 3. On the season, he now has 8.5 sacks, tied for fourth in the NFL.

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New England Patriots linebacker Matt Judon is first with 11.5, San Francisco 49ers defensive end Nick Bosa and Minnesota Vikings edge rusher Za’Darius Smith are tied for second with 9.5 sacks. Baltimore’s Justin Houston also has 8.5.

Highsmith has already broke his career high in sacks — he had six in 2021 — despite facing significantly more double-team blocks while T.J. Watt missed seven weeks with an injury. This year, Highsmith has 36 tackles, eight tackles for loss, three forced fumbles and a pass defended.

Indianapolis Colts running back Jonathan Taylor was the AFC Offensive Player of the Week. Tennessee Titans punter Ryan Stonehouse was the conference Special Teams Player of the Week. Minnesota Vikings wide receiver Justin Jefferson, Tampa Bay Buccaneers linebacker Devin White and Washington Commanders kicker Joey Slye won the NFC Awards for Week 10.